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3 years ago

Why does chocolate melt faster when crushed?

Chemistry

2 answers:

SSSSS [86.1K]3 years ago

7 0

Answer:

It is because when a chocolate is crushed, it will be smaller pieces which exerts a larger surface area. So that it can easily take in heat to give out reaction.

solong [7]3 years ago

5 0

Less compassionate around the chocolate

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Chlorine dioxide is used as a disinfectant and bleaching agent. In water, it reacts to form chloric acid (HClO3),

Sliva [168]

All done no worries

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3 years ago

How many tons of zinc can be obtained from 976 tons of zinc ore that is 78.0% zinc carbonate?

kramer

First, calculating the amount of zinc carbonate by multiplying the given mass of the ore by the percent zinc carbonate.
mass of zinc carbonate = (976 tons)(0.78)
= 761.28 tons
Then, calculate for %Zn in zinc carbonate
%Zn = ((65.38) / (125.39)) x 100%
% Zn = 52.141%
tons of Zn from the sample,
mass of Zn = (761.28 tons)(0.52141)
= 396.94 tons Zn

6 0

3 years ago

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Give 3 examples of solutions, suspensions and colloids..

Vlad [161]

Some examples of solutions":" are salt water, rubbing alcohol, and sugar dissolved in water.

examples of suspensions are muddy water milk of magnesia Slaked lime for whitewashing.

examples of colloids are whipped cream, mayonnaise, milk,

5 0

2 years ago

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Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion= $\frac{Volume}{Time}$

Effusion rate of oxygen gas after time t = $E=\frac{4.64 mL}{t}$

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = $E'=\frac{3.12 mL}{t}$

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{E}{E'}=\sqrt{\frac{M'}{M}}$

$\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}$

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

8 0

3 years ago

What is the pressure of 1.20 moles of SO2 gas in a 4L container at 30 degrees celsius

geniusboy [140]

Let suppose the Gas is acting Ideally, Then According to Ideal Gas Equation,

P V = n R T
Solving for P,
P = n R T / V ----- (1)
Data Given;

Moles = n = 1.20 mol

Volume = V = 4 L

Temperature = T = 30 + 273 = 303 K

Gas Constant = R = 0.08206 atm.L.mol⁻¹.K⁻¹

Putting Values in Eq.1,

P = (1.20 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 303 K) ÷ 4 L

P = 7.45 atm

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3 years ago

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